This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
Let's solve these two problems. Problem 1: Plane Flight a) Find the bearing and distance of the plane from its starting point. Step 1: Break down each leg into North/South and East/West components. For a bearing θ (clockwise from North): North component = Distance × cos(θ) East component = Distance × sin(θ) Leg 1: 300 km on a bearing of 30° North₁ = 300 km × cos(30°) = 300 km × (√3 / 2) = 150√3 km East₁ = 300 km × sin(30°) = 300 km × (1/2) = 150 km Leg 2: 450 km on a bearing of 270° (due West) North₂ = 450 km × cos(270°) = 450 km × 0 = 0 km East₂ = 450 km × sin(270°) = 450 km × (-1) = -450 km (or 450 km West) Step 2: Calculate the total North/South and East/West displacement. Total North displacement (N_total) = North₁ + North₂ N_total = 150√3 km + 0 km = 150√3 km ≈ 259.81 km Total East displacement (E_total) = East₁ + East₂ E_total = 150 km + (-450 km) = -300 km (or 300 km West) Step 3: Calculate the total distance from the starting point. The total distance (D) is the hypotenuse of a right triangle formed by N_total and E_total. D = √(N_total² + E_total²) D = √((150√3)² + (-300)²) D = √( (22500 × 3) + 90000 ) D = √( 67500 + 90000 ) D = √(157500) D = 150√7 km D ≈ 396.9 km (to one decimal place) Step 4: Calculate the bearing from the starting point. The plane is North (N_total is positive) and West (E_total is negative) of the starting point. Let α be the angle measured from the North axis towards the West. tan(α) = |E_total| / |N_total| tan(α) = 300 / (150√3) = 2 / √3 α = arctan(2/√3) ≈ 49.1° The bearing is measured clockwise from North. Since the displacement is in the North-West quadrant, the bearing is 360° - α. Bearing = 360° - 49.1° = 310.9° b) Find how far north and how far west the plane from its starting point. From Step 2: The plane is 150√3 km (or approximately 259.8 km) North of its starting point. The plane is 300 km West of its starting point. Problem 2: Radio Mast Height Step 1: Identify the trigonometric relationship. We have a right-angled triangle where: The angle of elevation is 60°. The distance from the base (adjacent side) is 54 m. The height of the mast (opposite side) is what we need to find. The tangent function relates the opposite and adjacent sides: tan(angle) = Opposite / Adjacent. Step 2: Substitute the known values into the formula. tan(60°) = Height / 54 m Step 3: Solve for the height. Height = 54 m × tan(60°) Height = 54 m × √3 Height ≈ 54 m × 1.73205 Height ≈ 93.5287 m Step 4: Round to the nearest meter. Height ≈ 94 m That's 2 down. 3 left today — send the next one.